Download Dynamics (B) concept WS – honors ANSWERS

Dynamics (B) concept WS – honors
ANSWERS
1. Two blocks of the same mass are connected by a string and are being pulled across a frictionless surface by a
constant force, F.
F
m1
m2
a. Will the two blocks move with constant velocity? Explain.
No; they will accelerate since the force F is the only horizontal force acting on the twoblock system.
b. Will the tension force exerted by the connecting string be greater than, less than, or equal to the force F?
Explain.
It will be less than F. Both blocks have the same acceleration (since they are
connected). The force F is essentially accelerating both masses together. The string is
essentially accelerating only mass m1. Since ΣF = ma and m1 = m2, the force F must be
twice as large as the tension force in the string.
c.
Which block, if either, will have the greater total force acting upon it? Explain.
Since both blocks have the same mass and the same accelerations, Newton’s 2nd Law
(ΣF = ma) states that they would have the SAME amount of total force acting on them.
Look at it like this: It was established in part B that the force F was twice as large as the
tension in the string. For mass 1, the tension force is the ONLY horizontal force so ΣF =
T. Mass 2 has the force F pulling one way and the tension force pulling the other.
Therefore, ΣF = F – T = (2T) – T = T
2. The diagram below shows a box being slid across a level floor by a constant pulling force, F.
θ
a.
Is the normal force acting on the box equal to the weight of the box? Explain.
No; it is LESS than the weight of the box. Since the force F is pulling up a little bit, it is
helping the floor cancel out the gravitational force on the box. Therefore, FN = mg - Fy
b.
What will happen to the acceleration of the box if angle θ is increased? Explain.
The vertical component of F will increase, causing the normal force on the box to
decrease. This will cause the frictional force acting on the block to decrease, which
should allow for a larger acceleration of the box. However, as θ increases, the horizontal
component of F will decrease which means it isn’t being pulled across the floor as hard,
which should cause the acceleration of the box to decrease. In reality, the acceleration
increases until some critical value of θ is reached, after which the acceleration decreases.
c.
If force F pushes down on the box (instead of pulling up) at the same angle θ, how will the acceleration
change? Explain.
Pushing down on the box will cause the normal force to be larger than the weight of the
box (FN = mg + Fy), which will cause the frictional force on the box to increase. Thus, the
acceleration of the box should be less than if it is pulled at that angle.
3. The following items deal with the ideal Atwood machine illustrated below:
m1
m2
a.
If m1 = m2 , how will the system move? Explain.
The system will maintain a constant velocity (0 or otherwise) since the external forces
acting on it (m1g & m2g) are balanced.
b.
If the acceleration of the system is equal to 1/3 g, clockwise, how does m 1 compare to m2?
If the system accelerates clockwise then m2 must be larger than m1. Therefore:
ΣF = ma
m2g – m1g = (m2 + m1) a
(m2 – m1)g = (m2 + m1) (1/3 g)
m2 – m1 = 1/3 m2 + 1/3 m1
2/3 m2 = 4/3 m1
m2 = 2 m 1
c.
If m1 = 2 m2, which end of the string exerts the greater tension force? Explain.
The tension force is THE SAME at both ends of the string. That is the nature of tension.
(Thanks to Newton’s 3rd Law on a molecular level)
d.
Generally speaking, how does the tension in the string compare to the weights of m 1 and m2?
m1g > T > m2g
e.
If m1 = 3m2 and m1 is placed on a frictionless horizontal table (while m2 hangs off the edge), what will the
system do?
The system will accelerate (m1 towards the edge of the table and m2 down toward the
floor)
ΣF = ma
m2g = (m2 + m1) a
m2g = (m2 + 3m2) a
m2g = (4 m2) a
g=4a
¼g=a
The acceleration of the system will be equal to one-quarter the acceleration due to
gravity.
4. The diagram below shows a wood box on an inclined wood plank.
FN
Ff
θ
mg
a.
Identify the forces acting on the box (disregarding air resistance).
Gravitational force, Frictional force, Normal Force (see diagram above)
b.
If angle θ is increased, what effect would that have on the frictional force acting on the box? Explain.
Increasing θ will cause the normal force on the block to decrease (since the perpendicular
component of ‘mg’ will decrease). This will cause the frictional force to decrease since
Ff = µ FN
c.
If the mass of the box is decreased, what effect would that have on the acceleration of the box? Explain.
The acceleration of the box would NOT CHANGE.
ΣF = ma
mg(parallel) – Ff = ma
mg sinθ - µ mg cosθ = ma
Since there is an ‘m’ in every term of the equation, it cancels out. Therefore:
a = g sinθ - µg cos θ
5. A sign is suspended by two cables, as illustrated in the diagram below.
rope 1
a.
rope 2
Is the sign in translational equilibrium? Explain.
Since the sign is not accelerating, then the net force on it is 0. Therefore, it is in
translational equilibrium.
b.
Does each rope experience the same amount of tension?
Since the ropes make different angles to the horizontal, they would not experience the
same amount of tension. The tension in rope 2 should be greater than the tension in rope
1.
(The ropes are both acting to cancel out the gravitational force on the sign. Since rope 2
is pulling more vertically than rope 1, it is doing more of the “work”.
c.
If the weight of the sign increases, what happens to the tension in rope 1?
If the weight of the sign increases, the y-components of both tension forces must
increase to cancel out ‘mg’. Therefore, the tension in both ropes would INCREASE.
d.
If the horizontal angle made by rope 1 is increased, what happens to the tension in rope 2?
If rope 1 makes a greater angle to the horizontal, it is now pulling more vertically than it
was. It is now helping rope 2 cancel out ‘mg’ to a greater extent. Therefore, the tension
in rope 2 will DECREASE.